Which one of the following is not a basic solution of system of linear equation

x1 + 2x2 + x3 = 4

2x1 + x2 + 5x3 = 5

The correct answer is : option 1

Key Points

Concept of Basic Solution:

• A basic solution to a system of linear equations in matrix form (Ax = b) is obtained by setting (n - m) variables to zero, where:
  • n = total number of variables
  • m = number of equations
• Here, we have 2 equations (m = 2) and 3 variables (x₁, x₂, x₃), so:
  • We must set (3 - 2) = 1 variable to zero to get a basic solution.
• If all variables are non-zero, it is not a basic solution.

Now, let’s analyze each option:

 Option 1) x₁ = -1, x₂ = 2, x₃ = 1

• All variables are non-zero → Not a valid basic solution.
• Check equation 1: -1 + 2×2 + 1 = -1 + 4 + 1 = 4 ✔
• Check equation 2: 2×(-1) + 2 + 5×1 = -2 + 2 + 5 = 5 ✔
• This is NOT a basic solution because no variable is zero.

 Option 2) x₁ = 2, x₂ = 1, x₃ = 0

• x₃ = 0 (one variable is zero) → could be a basic solution.
• Equation 1: 2 + 2×1 + 0 = 4 ✔
• Equation 2: 2×2 + 1 + 5×0 = 4 + 1 = 5 ✔
• Valid basic solution.

 Option 3) x₁ = 5, x₂ = 0, x₃ = -1

• x₂ = 0 → One variable zero → Candidate for basic solution.
• Equation 1: 5 + 0 + (-1) = 4 ✔
• Equation 2: 2×5 + 0 + 5×(-1) = 10 - 5 = 5 ✔
• Valid basic solution.

 Option 4) x₂ = 5/3, x₃ = 2/3

• Need to find x₁:
• Equation 1: x₁ + 2×(5/3) + (2/3) = 4 → x₁ + 10/3 + 2/3 = 4 → x₁ = 4 - 12/3 = 0
• x₁ = 0 → One variable is zero → Possible basic solution
• Equation 2: 2×0 + 5/3 + 5×(2/3) = 5/3 + 10/3 = 15/3 = 5 ✔
• Valid basic solution.

✅ Final Answer:

Option 1) x₁ = -1, x₂ = 2, x₃ = 1 is NOT a basic solution because all variables are non-zero.